Question #a0129

I got #"0.4072 atm"#.

This implies that by knowing

#"0.1026 atm" harr 20.00^@ "C"#
#"0.4072 atm" harr 54.00^@ "C"#
#"1.000 atm" harr 80.74^@ "C"#

the slope of the liquid-vapor coexistence curve on the phase diagram of cyclohexane is positive. That's normal, as most if not all liquids expand when they vaporize...

The Clapeyron equation approximates the slope of a two-phase equilibrium line:

#(dP)/(dT) = (DeltabarH)/(TDeltabarV)#

where:

• #DeltabarH# is the change in molar enthalpy.
• #T# is the temperature in #"K"#.
• #barV = V/n# is the molar volume of the substance.

In considering boiling point we examine the liquid-vapor coexistence curve (#bar(TC)#).

So, assuming the vapor is ideal, #barV = (RT)/P#, and:

#(dP)/(dT) = (DeltabarH cdot P)/(RT^2)#

Therefore, we can then write the derivative form of the Clausius-Clapeyron equation:

#1/P(dP)/(dT) = (DeltabarH_"vap")/(RT^2) = (dlnP)/(dT)#

You may recognize this better once we separate the variables and integrate.

#int_(P_1)^(P_2)1/PdP = int_(T_1)^(T_2)(DeltabarH_"vap")/(RT^2)dT#

We assume the change in molar enthalpy due to vaporization remains constant in the small temperature range (i.e. that this curve is straight!) to obtain the Clausius-Clapeyron Equation:

#=> color(green)barul(|stackrel(" ")(" "ln(P_2/P_1) = -(DeltabarH_"vap")/R [1/T_2 - 1/T_1]" ")|)#

and if you were to plot the #ln# of a set of pressure data points on the #y# axis and a set of corresponding inverse boiling points #1/T# on the #x# axis, you would obtain a slope of #-(DeltabarH_"vap")/R#.

And now our equation relates vapor pressures #P_i# with boiling temperatures #T_i#, using #DeltabarH_"vap"# as part of our slope.

Now, you were given:

• The normal boiling point of #80.74^@ "C"# occurs implicitly at #"1.000 atm"# vapor pressure.
• At #20.00^@ "C"#, the vapor pressure is #"0.1026 atm"#.

At this point you can then solve for the enthalpy of vaporization.

#ln("1.000 atm"/"0.1026 atm") = -(DeltabarH_"vap")/("0.008314472 kJ/mol"cdot"K") [1/(80.74 + "273.15 K") - 1/(20.00 + "273.15 K")]#

Therefore:

#DeltabarH_"vap" = -ln(9.747) cdot "0.008314472 kJ/mol"cdot"K" cdot [-5.855 xx 10^(-4) "K"^(-1)]^(-1)#

#=# #ul"32.33 kJ/mol"#

Apparently the actual value is around #"30.1 kJ/mol"#, so we're not that far off.

So now, if we want the vapor pressure at #54.00^@ "C"#, we can use the #20.00^@ "C"# as a new #T_1# reference point and solve for #P_2# using #T_2 = 54.00^@ "C"#. That is, we found what we needed to get to the slope, and now we are re-using it to find a single data point.

#ln(P_2/"0.1026 atm") = -("32.33 kJ/mol")/("0.008314472 kJ/mol"cdot"K") [1/(54.00 + "273.15 K") - 1/(20.00+"273.15 K")]#

#= 1.379#

As a result, the vapor pressure at the intermediate temperature is:

#color(blue)(P_2) = "0.1026 atm" cdot e^(1.379) ~~ color(blue)ul("0.4072 atm")#